C Program for Array Rotation

A class stores five marks in their original order: 40, 35, 45, 35 and 50. The rotate an array left task needs a clear rule about order, duplicates and how many elements exist.

A rotation moves an endpoint around to the other end while preserving all values and their cyclic order.

Before this exercise, review array elements and bounds and function parameters. The worked program below applies those ideas to this specific problem.

Explain
Explain why rotate an array left uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Rotate the same original array one position right instead.

Rotate an Array Left: follow the data

Given data
A class stores five marks in their original order: 40, 35, 45, 35 and 50. The rotate an array left task needs a clear rule about order, duplicates and how many elements exist.
Rule
A rotation moves an endpoint around to the other end while preserving all values and their cyclic order.
Observed result
35 45 35 50 40

A guided model of this example. The illustration explains the process; run the complete C program to check its output.

Understand the operation

Save the first element before shifting left. A right rotation instead saves the last and shifts from right to left so an unread value is not overwritten.

The complete example knows its five-element bound. In a reusable function, pass an explicit element count with the array pointer; a pointer alone does not report that bound. This sample has two equal marks, which makes duplicate handling observable.

Read the complete C program

The input values are written into this complete C17 example so the calculation can be reproduced without an unavailable service. Change the named data, rebuild and compare the actual output with your prediction.

C17 · complete program

#include <stdio.h>
int main(void) {
  int marks[] = {40, 35, 45, 35, 50};
  int count = (int)(sizeof marks / sizeof marks[0]);
  int first = marks[0];
  for (int i = 0; i < count - 1; i++) marks[i] = marks[i + 1];
  marks[count - 1] = first;
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

Expected output

35 45 35 50 40

Run in C compiler →

Follow the values through the code

C17 · code used in this explanation

#include <stdio.h>
int main(void) {
  int marks[] = {40, 35, 45, 35, 50};
  int count = (int)(sizeof marks / sizeof marks[0]);
  int first = marks[0];
  for (int i = 0; i < count - 1; i++) marks[i] = marks[i + 1];
  marks[count - 1] = first;
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

The saved first value is 40; the other four move one position left, and 40 becomes the final element.

Compare the resulting output with the original sequence. A changed order and a changed set of values are different effects; decide which your operation promises.

Follow the worked calculation

Start with the sample
The saved first value is 40; the other four move one position left, and 40 becomes the final element.
Apply the rule
Compare the resulting output with the original sequence. A changed order and a changed set of values are different effects; decide which your operation promises.
3
Check the result
35 45 35 50 40
```
35 45 35 50 40
```

A guided model of this example. The illustration explains the process; run the complete C program to check its output.

Try a small variation

C17 · code used in this explanation

#include <stdio.h>
int main(void) {
  int marks[] = {40, 35, 45, 35, 50};
  int count = (int)(sizeof marks / sizeof marks[0]);
  int first = marks[0];
  for (int i = 0; i < count - 1; i++) marks[i] = marks[i + 1];
  marks[count - 1] = first;
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

Rotate the same original array one position right instead.

Worked answer

The complete answer keeps the five valid input elements and applies the stated alternative. Read the output against the original array rather than assume that every loop should sort, remove or overwrite values.

C17 · complete program

#include <stdio.h>
int main(void) {
  int marks[] = {40, 35, 45, 35, 50};
  int count = (int)(sizeof marks / sizeof marks[0]);
  int last = marks[count - 1];
  for (int i = count - 1; i > 0; i--) marks[i] = marks[i - 1];
  marks[0] = last;
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

Expected output

50 40 35 45 35

Run in C compiler →

Recognize the important boundaries

An empty array needs a separate no-operation case before endpoint access. For rotation by k, define direction and reduce k modulo a nonzero count.

The sample marks and their sums fit int. General input requires both valid array bounds and an arithmetic-range contract.

Choose the answer, then compare the reasoning for every option.

Use the idea independently

A rotation moves an endpoint around to the other end while preserving all values and their cyclic order.

Continue with array indexes and bounds.

Continue with pointer and count contracts.

Language reference: WG14 C11 committee draft. These examples use the C17-compatible subset of the explained language features.

Reaching the recap and completing the quizzes are separate milestones.

Next lesson: C Program to Remove Duplicate Elements From a Sorted Array →

C Program for Array Rotation